Question

Doar exercițiile 29, 32 şi 34, vă rog!

Answer 1

$29) {3}^{ {x}^{2} - 2x + 5 } = 81$

${3}^{ {x}^{2} - 2x + 5 } = {3}^{4}$

${x}^{2} - 2x + 5 = 4$

${x}^{2} - 2x + 5 - 4 = 0$

${x}^{2} - 2x + 1 = 0$

${(x - 1)}^{2} = 0$

$x - 1 = 0 = > x = 1 \: \in \: \mathbb{R}$

$32) {2}^{ {x}^{2} + 3x + 4} = 4$

${2}^{ {x}^{2} + 3x + 4 } = {2}^{2}$

${x}^{2} + 3x + 4 = 2$

${x}^{2} + 3x + 4 - 2 = 0$

${x}^{2} + 3x + 2 = 0$

${x}^{2} + 2x + x + 2 = 0$

$x(x + 2) + 1(x + 2) = 0$

$(x + 1)(x + 2) = 0$

$x + 1 = 0 = > x = - 1\:\in\:\mathbb{R}$

$x + 2 = 0 = > x = - 2\:\in\:\mathbb{R}$

$34) {5}^{ {x}^{2} + 4x + 7 } = 625$

${5}^{ {x}^{2} + 4x + 7} = {5}^{4}$

${x}^{2} + 4x + 7 = 4$

${x}^{2} + 4x + 7 - 4 = 0$

${x}^{2} + 4x + 3 = 0$

${x}^{2} + 3x + x + 3 = 0$

$x(x + 3) + 1(x + 3) = 0$

$(x + 1)(x + 3) = 0$

$x + 1 = 0 = > x = - 1\:\in\:\mathbb{R}$

$x + 3 = 0 = > x = - 3\:\in\:\mathbb{R}$